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Lecture 5

Lecture 5. Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which they take place. Lecture 5 – Thursday. Block 1: Mole Balances Block 2: Rate Laws Block 3: Stoichiometry

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Lecture 5

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  1. Lecture 5 Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which they take place.

  2. Lecture 5 – Thursday • Block 1: Mole Balances • Block 2: Rate Laws • Block 3: Stoichiometry • Stoichiometric Table: Flow • Definitions of Concentration: Flow • Gas Phase Volumetric Flow Rate • Calculate the Equilibrium Conversion Xe

  3. Review Lecture 2 ReactorMole Balances Summary in terms of conversion, X X Batch t CSTR X W PFR PBR

  4. Review Lecture 3 Algorithm How to find Step 1: Rate Law Step 2:Stoichiometry Step 3: Combine to get

  5. Review Lecture 3 Reaction Engineering Mole Balance Rate Laws Stoichiometry These topics build upon one another

  6. Flow System Stoichiometric Table Where: and Concentration – Flow System

  7. Stoichiometry Concentration Flow System: LiquidPhase Flow System: Flow LiquidPhase Liquid Systems etc.

  8. Liquid Systems If the rate of reactionwere thenwewouldhave This gives us

  9. StoichiometryforGas Phase Flow Systems Combining the compressibility factor equation of state with Z = Z0 Stoichiometry: Weobtain:

  10. StoichiometryforGas Phase Flow Systems Since , Using the same method,

  11. StoichiometryforGas Phase Flow Systems The total molar flow rate is: Substituting FT gives: Where

  12. For Gas Phase Flow Systems Concentration Flow System: Gas Phase Flow System:

  13. For Gas Phase Flow Systems If –rA=kCACB This gives us FA0/-rA X

  14. For Gas Phase Flow Systems

  15. Example: Calculating the equilibrium conversion (Xef) for gas phasereaction in a flow reactor Consider the following elementary reaction where KC=20 dm3/mol and CA0=0.2 mol/dm3. CalculateEquilibriumConversion or both a batch reactor (Xeb) and a flow reactor (Xef).

  16. Gas Flow Example (Xef) Rate Law: Solution:

  17. Gas Flow Example (Xef)

  18. Gas Flow Example (Xef) Stoichiometry: Gas isothermal T=T0 Gas isobaric P=P0

  19. Gas Flow Example (Xef) Pure A  yA0=1, CA0=yA0P0/RT0, CA0=P0/RT0 At equilibrium: -rA=0

  20. Gas Flow Example (Xef) Recall Flow: Batch:

  21. End of Lecture 5

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